3 Answers
3

Yes. It is a theorem of Cayley that the mapping $S \rightarrow (S-I)^{-1}(S+1)$ gives a correspondence between the set of $n\times n$ skew-symmetric matrices over $\mathbb{Q}$ and the set of $n\times n$ orthogonal matrices which do not have one as an eigenvalue. Since the mapping is nice, and rational skew-symmetric matrices are dense in the set of all skew-symmetric matrices, you have your result. For more, see the very nice paper by Liebeck and Osborne

That was my answer as well. I would also recommend Weyl's gem, The Classical Groups. Their Invariants and Representations. He has a whole section devoted to the Cayley parameterization. (Chap. II. Sec II)
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Liviu NicolaescuMar 2 '12 at 20:55

Sure. Consider matrices which fix $n-2$ of the standard basis vectors and describe a rotation in the plane spanned by the last two about an angle $\theta$ such that $\sin \theta, \cos \theta$ are both rational; these are dense in all such rotations, and all such rotations generate the orthogonal group, so the corresponding products (all of which are rational) are dense in the orthogonal group.

I should say yes. For this, I shall use the fact that in the unit sphere $\mathbb S^{d-1}$, the set of rational vectors is dense. I shall proceed by induction over $n$.

So let $A\in {\bf O}_n(\mathbb R)$ be given. Let $\vec v_1$ be its first column, an element of ${\mathbb S}^{n-1}$. We can choose a rational unit vector $\vec w_1$ arbitrarily close to $\vec v_1$. The first step is to construct a rational orthogonal matrix $B$ with first column $\vec w_1$. To this end we choose inductively rational unit vectors $\vec w_2,\ldots,\vec w_n$. This is possible because at each step, we may take a rational unit vector in the unit sphere of a "rational" subspace. Here, a subspace $F$ is rational if it admits a rational basis.

Now, let us form $A_1=B^{-1}A$. This is a orthogonal matrix, whose first column is arbitrarily close to $\vec e_1$. Hence its first line is close to $(1,0,\ldots,0)$ as well. Thus
$$A_1\sim\begin{pmatrix} 1 & 0^T \\\\ 0 & R \end{pmatrix}.$$
The matrix $R$ is arbitrarily close to ${\bf O}_{n-1}({\mathbb R})$. By the induction hypothesis, there exists a rational orthogonal matrix $Q$ arbitraly close to $R$. Then
$$B\begin{pmatrix} 1 & 0^T \\\\ 0 & Q \end{pmatrix}$$
is arbitrarily close to $A$.